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A point mass oscillates along the $x$-axis according to the law $\mathrm{x=x_0 \cos (\omega t-\pi / 4)}$. If the acceleration of the particle is written as
$\mathrm{a=A \cos (\omega t+\delta)}$
Options:
$\mathrm{a=A \cos (\omega t+\delta)}$
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Verified Answer
The correct answer is:
$\mathrm{A=x_0 \omega^2, \delta=3 \pi / 4}$
$\mathrm{A=x_0 \omega^2, \delta=3 \pi / 4}$
$\mathrm{v=-x_0 \omega \sin (\omega t-\pi / 4)}$
$\mathrm{a=-x_0 \omega^2 \cos \left(\omega t+\pi-\frac{\pi}{4}\right)}$
$\mathrm{a=A \cos (\omega t+\delta)}$
$\mathrm{A=x_0 \omega^2 ; \quad \delta=\frac{3 \pi}{4}}$
$\mathrm{a=-x_0 \omega^2 \cos \left(\omega t+\pi-\frac{\pi}{4}\right)}$
$\mathrm{a=A \cos (\omega t+\delta)}$
$\mathrm{A=x_0 \omega^2 ; \quad \delta=\frac{3 \pi}{4}}$
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